Đáp án:
$\begin{array}{l}
a)Dkxd:x - 1 \ne 0 \Rightarrow x \ne 1\\
\dfrac{{3 - 5x}}{{x - 1}} = 0\\
\Rightarrow 3 - 5x = 0\\
\Rightarrow x = \dfrac{3}{5}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{3}{5}\\
b)\left| {2x - 3} \right| + \left| { - 3x - 1} \right| = 0\\
Do:\left\{ \begin{array}{l}
\left| {2x - 3} \right| \ge 0\\
\left| { - 3x - 1} \right| \ge 0
\end{array} \right.\\
\Rightarrow \left| {2x - 3} \right| = \left| { - 3x - 1} \right| = 0\\
\Rightarrow \left\{ \begin{array}{l}
2x - 3 = 0\\
- 3x - 1 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{1}{3}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,ko\,có\,x\,thỏa\,mãn.\\
3)\dfrac{x}{{ - 2x + 3}}\,ko\,xac\,dinh\\
\Rightarrow - 2x + 3 = 0\\
\Rightarrow x = \dfrac{3}{2}
\end{array}$
